Question

A Carnot engine, having an efficiency of `eta=1//10` as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is
A. `99J`
B. `90J`
C. `1J`
D. `100J`

Answer 1

Correct Answer - B
Let `eta` is the efficiency of heat engine and `beta` is the corresponding coefficient of performance of a refrigerator working between the same temperature.
`beta = (1)/(eta) -1 = 10-1 = 9 "Also" beta=("Output")/("Input")`
=`"(Energy absorbed from the reservoir at lower temperature")/("Work done on the system")`
`beta = 9 = ("Energy absorbed")/(10J)`
So , energy absorbed = `90J`